ZOJ 2852 Deck of Cards(DP)

题目连接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1852

题意太繁琐就不讲了.

设dp[q][i][j][k]表示前i张牌,每个卡槽的分值各为i,j,k时候的最能的多金钱数.

dp[0][0][0][0]=0,其他设为-1.

用了滚动数组优化空间,具体的转移方程看程序吧.

#include <cstdio>
#include <algorithm>
#include <memory.h>
#define JOKER 21
#define FACE 10
using namespace std;
const int MAX = 101;


int convert_to_val(char c){
	if(c == ‘F‘)return JOKER;
	else if(c == ‘A‘)return 1;
	else if(c >= ‘2‘ && c <= ‘9‘)return c - ‘0‘;
	else if(c == ‘T‘ || c == ‘J‘ || c == ‘Q‘ || c == ‘K‘)return FACE;
}

int main(int argc, char const *argv[]){
	int dp[2][31][31][31];
	int N;
	while(scanf("%d", &N) && N){
		int ans = -1;
		memset(dp, -1, sizeof(dp));
		dp[0][0][0][0] = 0;
		int f = 0;
		for(int p = 0; p <= N; ++p, f ^= 1){
			int nf = f ^ 1;
			char card[2];
			if(p != N)
				scanf("%s", card);
			else
				card[0] = ‘F‘;
			int val = convert_to_val(card[0]);
			for(int i = 0; i < 31; ++i){
				for(int j = 0; j < 31; ++j){
					for(int k = 0; k < 31; ++k){
						if(dp[f][i][j][k] == -1)continue;
						ans = max(ans, dp[f][i][j][k]);
						if(val == JOKER){//joker case
							if(i <= 20)
								dp[nf][0][j][k] = max(dp[nf][0][j][k], dp[f][i][j][k] + 50 + 100);
							if(j <= 20)
								dp[nf][i][0][k] = max(dp[nf][i][0][k], dp[f][i][j][k] + 50 + 200);
							if(k <= 20)
								dp[nf][i][j][0] = max(dp[nf][i][j][0], dp[f][i][j][k] + 50 + 300);
						}else{
							int nv = i + val;
							if(nv == 21){//reach 21
								dp[nf][0][j][k] = max(dp[nf][0][j][k], dp[f][i][j][k] + 50 + 100);
							}else if(i < 21){
								dp[nf][nv][j][k] = max(dp[nf][nv][j][k], dp[f][i][j][k] + 50);
							}
							
							nv = j + val;
							if(nv == 21){//reach 21
								dp[nf][i][0][k] = max(dp[nf][i][0][k], dp[f][i][j][k] + 50 + 200);
							}else if(j < 21){
								dp[nf][i][nv][k] = max(dp[nf][i][nv][k], dp[f][i][j][k] + 50);
							}

							nv = k + val;
							if(nv == 21){//reach 21
								dp[nf][i][j][0] = max(dp[nf][i][j][0], dp[f][i][j][k] + 50 + 300);
							}else if(k < 21){
								dp[nf][i][j][nv] = max(dp[nf][i][j][nv], dp[f][i][j][k] + 50);
							}
						}
					}
				}
			}
			memset(dp[f], -1, sizeof(dp[f]));
		}

		printf("%d\n", ans);
	}
	return 0;
}

ZOJ 2852 Deck of Cards(DP),布布扣,bubuko.com

ZOJ 2852 Deck of Cards(DP)

原文：http://blog.csdn.net/zxjcarrot/article/details/23039361